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Saturday, May 19, 2012

Class XI, CHEMISTRY, "Energetics Of Chemical Reaction"

ThermodynamicsDefinitionIt is branch of chemistry which deals with the heat energy change during a chemical reaction.Types of Thermochemical Reactions
Thermo-chemical reactions are of two types.
1. Exothermic Reactions
2. Endothermic Reactions1. Exothermic Reaction
A chemical reaction in which heat energy is evolved with the formation of product is known as Exothermic Reaction.
An exothermic process is generally represented as
Reactants —-> Products + Heat2. Endothermic Reaction
A chemical reaction in which heat energy is absorbed during the formation of product is known as endothermic reaction.
Endothermic reaction is generally represented as
Reactants + Heat —-> ProductsThermodynamic Terms1. System
Any real or imaginary portion of the universe which is under consideration is called system.2. Surroundings
All the remaining portion of the universe which is present around a system is called surroundings.3. State
The state of a system is described by the properties such as
temperature, pressure and volume when a system undergoes a change of
state, it means that the final description of the system is different
from the initial description of temperature, pressure or volume.Properties of System
The properties of a system may be divided into two main types.1. Intensive Properties
Those properties which are independent of the quantity of matter are called intensive properties.
e.g. melting point, boiling point, density, viscosity, surface, tension, refractive index etc.2. Extensive Properties
Those properties which depends upon the quantity of matter are called extensive properties.
e.g. mass, volume, enthalpy, entropy etc.First Law of Thermodynamics
This law was given by Helmheltz in 1847. According to this law
Energy can neither be created nor destroyed but it can be changed from one form to another.
In other words the total energy of a system and surroundings must remain constant.Mathematical Derivation of First Law of Thermodynamics
Consider a gas is present in a cylinder which contain a frictionless piston as shown.
Diagram Coming Soon
Let a quantity of heat q is provided to the system from the
surrounding. Suppose the internal energy of the system is E1 and after
absorption of q amount of heat it changes to E2. Due to the increase of
this internal energy the collisions offered by the molecules also
increases or in other words the internal pressure of the system is
increased after the addition of q amount of heat. With the increase of
internal pressure the piston of the cylinder moves in the upward
direction to maintain the pressure constant so a work is also done by
the system.
Therefore if we apply first law of thermodynamics on this system we can write
q = E2 – E1 + W
OR
q = ΔE + W
OR
ΔE = q – W
This is the mathematical representation of first law of thermodynamics.Pressure – Volume Work
Consider a cylinder of a gas which contain a frictionless and weightless
piston, as shown above. Let the area of cross-section of the piston = a
Pressure on the piston = P
The initial volume of the gases = V1
And the final volume of the gases = V2
The distance through which piston moves = 1
So the change in volume = ΔV = V2 – V1
OR ΔV = a x 1
The word done by the system W = force x distance
W = Pressure x area x distance
W = P x a x 1
W = P Δ V
By substituting the value of work the first law of thermodynamics may be written as
q = ΔE + P ΔV
The absorption or evolution of heat during chemical reaction may take place in two ways.1. Process at Constant Volume
Let qv be the amount of heat absorbed at constant volume.
According to first law qv = ΔE + P ΔV
But for constant volume ΔV = O
Therefore,
P ΔV = P x O = O
So,
qv = ΔE + 0
Or
qv = ΔE
Thus in the process carried at constant volume the heat absorbed or evolved is equal to the energy ΔE.2. Process at Constant Pressure
Let qp is the amount of heat energy provided to a system at constant
pressure. Due to this addition of heat the internal energy of the gas is
increased from E1 to E2 and volume is changed from V1 to V2, so
according to first law.
qp = E2 – E1 + P(V2 – V1)
Or
qp = E2 – E1 + PV2 – PV1
Or
qp = E2 + PV2- E1 – PV1
Or
qp = (E2 + PV2) – (E1 – PV1)
But we known that
H = E + PV
So
E1 + PV1 = H1
And
E2 + PV2 = H2
Therefore the above equation may be written as
qp = H2 – H1
Or
qp = Δ H
This relation indicates that the amount of heat absorbed at constant pressure is used in the enthalpy change.Sign of ΔH
ΔH represent the change of enthalpy. It is a characteristic property of a
system which depends upon the initial and final state of the system.
For all exothermic processes ΔH is negative and for all endothermic reactions ΔH is positive.Thermochemistry
It is a branch of chemistry which deals with the measurement of heat evolved or absorbed during a chemical reaction.
The unit of heat energy which are generally used are Calorie and kilo Calorie or Joules and kilo Joules.
1 Cal = 4.184 J
OR
1 Joule = 0.239 CalHess’s Law of Constant Heat SummationStatement
If a chemical reaction is completed in a single step or in several steps
the total enthalpy change for the reaction is always constant.
OR
The amount of heat absorbed or evolved during a chemical reaction must
be independent of the particular manner in which the reaction takes
place.Explanation
Suppose in a chemical reactant A changes to the product D in a single step with the enthalpy change ΔH
Diagram Coming Soon
This reaction may proceed through different intermediate stages i.e., A
first changes to B with enthalpy change ΔH1 then B changes to C with
enthalpy change ΔH2 and finally C changes to D with enthalpy ΔH3.
According to Hess’s law
ΔH = ΔH1 + ΔH2 + ΔH3Verification of Hess’s Law
When CO2 reacts with excess of NaOH sodium carbonate is formed with the
enthalpy change of 90 kJ/mole. This reaction may take place in two steps
via sodium bicarbonate.
In the first step for the formation of NaHCO3 the enthalpy change is -49
kJ/mole and in the second step the enthalpy change is -41 kJ/mole.
According to Hess’s Law
ΔH = ΔH1 + ΔH2
ΔH = -41 -49 = -90 kJ/mole
The total enthalpy change when the reaction is completed in a single
step is -90 kJ/mole which is equal to the enthalpy change when the
reaction is completed into two steps. Thus the Hess’s law is verified
from this example